= Solution
Let $d_{\mu,A}$ and $d_{\mu,B}$ be corresponding real 4d orbitals on the upper and lower metals, with identical global-axis phases. Their <symmetry-adapted linear combinations> are
$$
\phi_{\mu,\pm}=\frac{d_{\mu,A}\pm d_{\mu,B}}{\sqrt2},
$$
with the normalization shown in the negligible-overlap approximation; including overlap changes the denominator but not the symmetry. Inversion exchanges the centres and each local d orbital has even intrinsic parity, so the plus combinations are gerade and the minus combinations ungerade.
$$
\begin{array}{c|cc|cc}
\text{metal orbital}&\phi_+&\phi_-&\text{Mo--Mo bonding}&\text{Mo--Mo antibonding}\\\hline
d_{z^2}&a_{1g}&a_{2u}&a_{1g}\ (\sigma)&a_{2u}\ (\sigma^*)\\
(d_{xz},d_{yz})&e_g&e_u&e_u\ (\pi)&e_g\ (\pi^*)\\
d_{xy}&b_{2g}&b_{1u}&b_{2g}\ (\delta)&b_{1u}\ (\delta^*)\\
d_{x^2-y^2}&b_{1g}&b_{2u}&b_{1g}\ (\delta)&b_{2u}\ (\delta^*)
\end{array}
$$
The reversal of bonding phase for $d_{xz},d_{yz}$ is important: the facing lobes have opposite signs when the two atomic orbitals share the same global phase convention, so their difference is bonding. Each $e$ irrep is two-dimensional. Thus
$$
\boxed{\Gamma_{10d}=A_{1g}+A_{2u}+B_{1g}+B_{2u}+B_{2g}+B_{1u}+E_g+E_u,}
$$
with total dimension ten. Rotating $x,y$ by $45^\circ$ interchanges the names of the two delta families; all labels here retain the chloride-aligned axes fixed above.
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